homework 02 – HOFFMA, DAVID – Due: Jan 28 2008, 4:00 am
1
Question 1, chap 23, sect 2.
part 1 of 3
10 points
A uniformly charged circular arc AB of ra
dius
R
is shown in the figure.
It covers a
quarter of a circle and it is located in the sec
ond quadrant. The total charge on the arc is
Q >
0.
The
value
of
the
Coloumb
constant
is
8
.
98755
×
10
9
N
·
m
2
/
C
2
.
x
y
+
+
+
+
+
+
+
+
+
Δ
θ
θ
R
x
y
I
II
III
IV
B
A
O
Δ
s
≡
R
Δ
θ
What is the direction of the electric field
vector
vector
E
at the origin, due to the charge
distribution?
1.
in quadrant III
2.
along the positive
x
axis
3.
along the negative
y
axis
4.
in quadrant IV
correct
5.
along the negative
x
axis
6.
along the positive
y
axis
7.
in quadrant I
8.
in quadrant II
Explanation:
The electric field for a positive charge is
directed away from it. In this case, the electric
field generated by each Δ
q
will be directed
into quadrant IV, so the total electric field
will be in the same quadrant.
Question 2, chap 23, sect 2.
part 2 of 3
10 points
Find
E
x
, the
x
component of the electric
field at the origin due to the full arc length for
the case, where
Q
= 2
.
5
μ
C and
R
= 1
.
42 m.
Correct answer: 7093
.
9
N
/
C (tolerance
±
1
%).
Explanation:
Let :
Q
= 2
.
5
μ
C = 2
.
5
×
10
−
6
C
,
R
= 1
.
42 m
,
and
k
e
= 8
.
98755
×
10
9
N
·
m
2
/
C
2
.
E
x
can be found by integrating the contri
butions of all the Δ
q
’s in the arc.
Using
Δ
q
=
Δ
θ
parenleftBig
π
2
parenrightBig
and taking the
x
component, then
adding the force component from each charge
Δ
q
at angle
θ
, we have
E
x
=
integraldisplay
π/
2
0
k
e
Q
R
2
2
π
cos
θ dθ
=
2
k
e
Q
π R
2
=
2 (8
.
98755
×
10
9
N
·
m
2
/
C
2
)
π
(1
.
42 m)
2
×
(2
.
5
×
10
−
6
C)
=
7093
.
9 N
/
C
.
Question 3, chap 23, sect 2.
part 3 of 3
10 points
If at
O
we have
E
x
=
C
, then what is the
magnitude of the full force
F
(not just the
x
component) on an electron at this point?
1.
F
=
C
3
e
2.
F
=
2
C
3
e
3.
F
=
3
e C
2
4.
F
=
C
e
√
2
5.
F
= 2
e C
6.
F
=
C
2
e
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homework 02 – HOFFMA, DAVID – Due: Jan 28 2008, 4:00 am
2
7.
F
=
e C
8.
F
= 3
e C
9.
F
= 4
e C
10.
F
=
√
2
e C
correct
Explanation:
From the symmetry of the charge distri
bution, it can be seen that
E
y
=
E
x
=
C
,
so
E
x
E
y
E
E
=
radicalBig
E
2
x
+
E
2
y
=
√
2
C .
The force on an electron is then
F
=
e E
=
√
2
e C
.
Question 4, chap 23, sect 2.
part 1 of 4
10 points
Consider a disk of radius 2
.
5 cm, having a
uniformly distributed charge of +4
.
2
μ
C.
The
value
of
the
Coulomb
constant
is
8
.
98755
×
10
9
N
·
m
2
/
C
2
.
Compute the magnitude of the electric field
at a point on the axis and 3
.
6 mm from the
center.
Correct answer: 1
.
03576
×
10
8
N
/
C (tolerance
±
1 %).
Explanation:
Let :
R
= 2
.
5 cm = 0
.
025 m
,
k
e
= 8
.
98755
×
10
9
N
·
m
2
/
C
2
,
Q
= 4
.
2
μ
C = 4
.
2
×
10
−
6
C
,
and
x
= 3
.
6 mm = 0
.
0036 m
.
The field at the distance
x
along the axis of
a disk with radius
R
is given by
E
= 2
π k
e
σ
parenleftbigg
1
−
x
√
x
2
+
R
2
parenrightbigg
,
where the surface charge density
σ
is
σ
=
Q
π R
2
=
4
.
2
×
10
−
6
C
π
(0
.
025 m)
2
= 0
.
00213904 C
/
m
2
.
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 Spring '08
 Turner
 Charge, Electrostatics, Work, Electric charge, DAVID

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